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uniform continuity

 
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Nov18-08, 10:25 PM   #1
 

uniform continuity


1. The problem statement, all variables and given/known data
suppose f and g are uniformly continuous functions on X

and f and g are bounded on X, show f*g is uniformly continuous.


3. The attempt at a solution

I know that if they are not bounded then they may not be uniformly continuous. ie x^2
and also if only one is bounded they are not necessarily uniformly continuous.

not sure what to do if they are both bounded
 
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Nov19-08, 06:04 AM   #2
 
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Quote by CarmineCortez View Post
1. The problem statement, all variables and given/known data
suppose f and g are uniformly continuous functions on X

and f and g are bounded on X, show f*g is uniformly continuous.


3. The attempt at a solution

I know that if they are not bounded then they may not be uniformly continuous. ie x^2
and also if only one is bounded they are not necessarily uniformly continuous.
No, you don't "know" that. In fact, here, you are told that they are both uniformly continuous.

not sure what to do if they are both bounded
Are you saying you could do this if only one were bounded? Do you understand what it is you are asked to prove?
 
Nov19-08, 08:13 AM   #3
 
uniform continuity is intuitively a bit like saying the function doesn't have an infinite slope anywhere...
And if they are both bounded, is their product too bounded?
 
Nov19-08, 09:48 AM   #4

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uniform continuity


Get your epsilons and deltas out. You want to show |f(x)g(x)-f(y)g(y)| can be made uniformly small if |x-y| is small. Hint: |f(x)g(x)-f(y)g(y)|=|f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)|. |f(x)-f(y)| and |g(x)-g(y)| can be made small since they are uniformly continuous. Do you see why f and g need to be bounded? Use epsilons and deltas to make the meaning of 'small' precise.
 
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